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3 years ago

50 POINTS If m∠1 = 53°, what is m∠4? A. 53° B. 43° C. 37° D. 27°

Mathematics

2 answers:

Debora [2.8K]3 years ago

8 0

Answer:

43.

Step-by-step explanation:

∠1 and ∠4 are vertical angles.

Vertical angles are equal.

Therefore,

m∠1 = m∠4 = 43°

Delicious77 [7]3 years ago

6 0

Ok so the answer is b

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Please help me with the below question.

VMariaS [17]

By letting

$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

we get derivatives

$y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}$

$y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}$

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

$5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0$

Examine the lowest degree term $\left(x^{r-1}\right)$ , which gives rise to the indicial equation,

$5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0$

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients $c_k$ is

$(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}$

so that with r = 4/5, the coefficients are governed by

$c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}$

c) Starting with $c_0=1$ , we find

$c_1 = -\dfrac{c_0}5 = -\dfrac15$

$c_2 = -\dfrac{c_1}{10} = \dfrac1{50}$

so that the first three terms of the solution are

$\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}$

4 0

2 years ago

klasskru [66]

Answer:

2. y - 2 = -1/6(x + 10)

3. y - 1 = -1/6(x + 4)

5. y = -1/6x + 1/3

Step-by-step explanation:

hope this helps!

8 0

2 years ago

Write an equation that models the sequence 6, 12, 24, 48... A) y = 6x B) y = 2x C) y = 6x2 D) y = 6(2x-1)

monitta

The answer is B) y=2x

8 0

4 years ago

Read 2 more answers

What is the sum of 20x^2-10x-30

ivann1987 [24]

Answer:

The sum of the roots is 0.5

Step-by-step explanation:

The correct question is

What is the sum of the roots of 20x^2-10x-30

we know that

In a quadratic equation of the form

$ax^{2} +bx+c=0$

The sum of the roots is equal to

$-\frac{b} {a}$

in this problem we have

$20x^{2} -10x-30=0$

$a=20\\b=-10\\c=-30$

substitute

$-\frac{(-10)} {20}=0.5$

Verify

Find the roots of the quadratic equation

The formula to solve a quadratic equation is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

$a=20\\b=-10\\c=-30$

substitute

$x=\frac{-(-10)\pm\sqrt{-10^{2}-4(20)(-30)}} {2(20)}$

$x=\frac{10\pm\sqrt{2,500}} {40}$

$x=\frac{10\pm50} {40}$

$x=\frac{10+50} {40}=1.5$

$x=\frac{10-50} {40}=-1$

The roots are x=-1 and x=1.5

The sum of the roots are

$-1+1.5=0.5$ ----> is ok

5 0

4 years ago

Anna said that the product of 78⋅112=7278·112=72.

sergejj [24]

Because she randomly added a 72 before the 78

3 0

3 years ago

Read 2 more answers